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Defined in header <type_traits>
template< class >
class result_of;
(1) (since C++11)
template< class F, class... ArgTypes >
class result_of<F(ArgTypes...)>;
(2) (since C++11)

Deduces the return type of a function call expression at compile time.

F must be a callable type, reference to function, or reference to callable type. Invoking F with ArgTypes... must be a well-formed expression (since C++11)
F and all types in ArgTypes can be any complete type, array of unknown bound, or (cv-qualified) void (since C++14)


[edit] Member types

Member type Definition
type the return type of the Callable type F if invoked with the arguments ArgTypes.... Only defined if F can be called with the arguments ArgTypes... in unevaluated context. (since C++14)

[edit] Helper types

template< class T >
using result_of_t = typename result_of<T>::type;
(since C++14)

[edit] Possible implementation

namespace detail {
template <class F, class... Args>
inline auto INVOKE(F&& f, Args&&... args) ->
    decltype(forward<F>(f)(forward<Args>(args)...)) {
      return forward<F>(f)(forward<Args>(args)...);
template <class Base, class T, class Derived>
inline auto INVOKE(T Base::*pmd, Derived&& ref) ->
    decltype(forward<Derived>(ref).*pmd) {
      return forward<Derived>(ref).*pmd;
template <class PMD, class Pointer>
inline auto INVOKE(PMD&& pmd, Pointer&& ptr) ->
    decltype((*forward<Pointer>(ptr)).*forward<PMD>(pmd)) {
      return (*forward<Pointer>(ptr)).*forward<PMD>(pmd);
template <class Base, class T, class Derived, class... Args>
inline auto INVOKE(T Base::*pmf, Derived&& ref, Args&&... args) ->
    decltype((forward<Derived>(ref).*pmf)(forward<Args>(args)...)) {
      return (forward<Derived>(ref).*pmf)(forward<Args>(args)...);
template <class PMF, class Pointer, class... Args>
inline auto INVOKE(PMF&& pmf, Pointer&& ptr, Args&&... args) ->
    decltype(((*forward<Pointer>(ptr)).*forward<PMF>(pmf))(forward<Args>(args)...)) {
      return ((*forward<Pointer>(ptr)).*forward<PMF>(pmf))(forward<Args>(args)...);
} // namespace detail
// Minimal C++11 implementation:
template <class> struct result_of;
template <class F, class... ArgTypes>
struct result_of<F(ArgTypes...)> {
    using type = decltype(detail::INVOKE(std::declval<F>(), std::declval<ArgTypes>()...));
// Conforming C++14 implementation (is also a valid C++11 implementation):
namespace detail {
template <typename, typename = void>
struct result_of {};
template <typename F, typename...Args>
struct result_of<F(Args...),
                 decltype(void(detail::INVOKE(std::declval<F>(), std::declval<Args>()...)))> {
    using type = decltype(detail::INVOKE(std::declval<F>(), std::declval<Args>()...));
} // namespace detail
template <class T> struct result_of : detail::result_of<T> {};

[edit] Notes

As formulated in C++11, std::result_of would fail to compile when F(ArgTypes...) is ill-formed (e.g. when F is not a callable type at all). C++14 changes that to a SFINAE (when F is not callable, std::result_of<F(Args...)> simply doesn't have the type member).

[edit] Examples

std::result_of can be used to determine the result of invoking a functor, in particular if the result type is different for different sets of arguments:

#include <type_traits>
struct S {
    double operator()(char, int&);
    float operator()(int);
struct C {
    double Func(char, int&);
int main()
    // the result of invoking S with char and int& arguments is double
    std::result_of<S(char, int&)>::type d = 3.14; // d has type double
    static_assert(std::is_same<decltype(d), double>::value, "");
    // the result of invoking S with int argument is float
    std::result_of<S(int)>::type f = 3.14; // f has type float
    static_assert(std::is_same<decltype(f), float>::value, "");
    // result_of can be used with a pointer to member function as follows
    std::result_of<decltype(&C::Func)(C, char, int&)>::type g = 3.14;
    static_assert(std::is_same<decltype(g), double>::value, "");

This program takes advantage of the C++14 changes to overload two functions on the "callability" of a template parameter:

#include <type_traits>
#include <iostream>
template<class T>
typename std::result_of<T(int)>::type
f(T& t)
    std::cout << "overload of f for callable T\n";
    return t(0);
template<class T, class U>
int f(U u)
    std::cout << "overload of f for non-callable T\n";
    return u;
struct S {};
int main() {
  f<S>(1); // fails to compile in C++11, calls the non-callable overload in C++14


overload of f for non-callable T

[edit] See also

obtains the type of expression in unevaluated context
(function template) [edit]